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Unformatted text preview: Problem 4.3. (Waste heat from a power plant.) (a) An eiﬁciency of 40% means that the other 60% of the energy consumed ends up as waste heat. That’s 1.5 times as much as the amount that ends up as work. More
generally, by the deﬁnition of efﬁciency and the ﬁrst law, W W e:—~ Qh :Q8+W’ so the waste heat is 6 Q6 = We — 1) = 1.5W 2 1.5 GW. (b) In one second, the waste heat dumped t0 the river is 1.5 x 109 J , and this heat is spread among 105 kg of water, so each kilogram gets 15 kJ. With a heat capacity of 4186 J/°C,
the water’s temperature increases by AT : Q/ C = 15000 J /4186 J/°C = 36°C. (c) rI‘he latent heat to evaporate water is 2260 J/g (at 100°C). At room temperature
it’s about 8% more, as mentioned in Problem 1.54 and Figure 5.11; so I’ll take L =
2400 J/g. The total amount of water that must evaporate each second is then 1.5 x 109 J = 6 105 : 6 0 k .
2400 J /g X g 0 g
That’s only 0.6 m3, or only 0.6% of the water in the river. Problem 4.4. (Engine driven by the ocean’s thermal gradient.) (a) Converting the temperatures to the kelvin scale, we get a maximum possible eﬂiciency Of T 277K
=1a—C: ———~=0.{)6
6 Th 1 295K 1’ or about 6%. (b) A rigorous calculation of the absolute minimum amount of water that we must process
is not easy. As the engine extracts heat from the warm water, the water’s temperature decreases and therefore so does the eﬂiciency of the engine. To make a rough estimate,
hOWever, let’s suppose that we extract heat from the warm water until its temperature
drops by 9°C (half the temperature difference between the warm and cool water), and
similarly that we expel heat into the cool water until its temperature increases by
9°C. Then the average temperatures of the reservoirs are 290.5 K and 281.5 K, so the efﬁciency is only 281.5 The heat extracted from each kilogram of the warm water is 9 x 4186 J = 38 kJ, but
at 3.1% efﬁciency, this heat produces only 1.2 kJ of work. We need 109 J of work each
second, so the amount of water required is 109 J h = 8. 105 k
1200 J/kg 6 X g’ or about 900 cubic meters. Problem 4.6. (Carnot cycle optimized for pOWer.) (a) (b) (C) The entropy gained by the engine as heat ﬂows in is Qh/Thw, while the entropy lost by the engine as heat flows out is Qc/Tcw. Assuming that no other entropy is created
inside the engine, these two quantities must be equal: ﬂ : QC Qh
Thur Tow QC But the ratio Qh/Qc can also be expressed in terms of the four temperatures using the
rate equations on the bottom of page 126. In this ratio, K and At cancel out, leaving us with
Th _ Thu. Thw
Tam _ Tc Tow  The power output is the work done per unit time. The time required for the heat Qh
to ﬂow in is Thu;
TC'UJ Qh
At = ———~—,
K(T;1 — TM”)
and the total time for the cycle is just twice this, so
_ W th’QCK(Th—Thw)mK Tcw
Power—mme—Tm§( Thw)(Th—Thw)1 where in the second step I’ve used energy conservation (W = Qh — Q6) and in the ﬁnal step I’ve again used Qh/Qc = Thw/Tm. Now solving the result of part (a) for
Tm, gives ‘ TC'UJ _ TC
Thu) — 2Thw —' Th ’ so K T
Power 2 3(1— (Th — Tm). To ﬁnd the value of Th1” that maximizes the power, set the derivative equal to zero: __ dPower _ K 2Tc _ T,2
— 2Thw _ Th. 0—. ~__._W_T_T _
, diam 2 (2Thw—Th)2(h ’1‘“) (1 _ 5 211m — Tam) — (21am — Th)? + TAZThw ~ Th.)
“ 2 W” (d) At this point the constant K/Z and the denominator are irrelevant, so the condition
is 0 : 2TcTh '_ 2113le _ + 4Th1hw _‘ + 2TcThw _ TcTh _ —4T,fw + 417,111” + Ten — T3. Applying the quadratic formula now gives 4T j: 4T 2~16Cr2—TCT i 1
Tm”: h ( h) 8 (h 0:11; :l:%\/TcTh:§(Thi TcTh)‘ The — sign in the :I: gives an unphysical solution, as you can see by considering the
case where TC is only slightly lower than Th; in this case the — sign would give Th,”
close to zero, whereas Thu, can’t be lower than TC. To ﬁnd the correSponding formula
for Tm, just plug into equation (1): TTh Tc  1(Th + x/TcTh) 1 T Th 1
Tcw=—c—~1—u—:—*~2—w~*=— C Tc =—Tc+v’TcT.
2Thm _ Th Th + V TCTh — Th 2 TcTh + 2( h) The efﬁciency in the case of maximum power is Qh Thw %(Th + vTcTh) VThU/q—L + x/Th) Th.
For the typical numbers given, this is
{298 K
8 ~ 1 w m —~v much closer to reality than the “ideal” efﬁciency of 1 — (298/873) = 66%. Of course, if the cost of fuel ever increases to the point where it outweighs the cost of power plant construction, it will become advantageous to design plants to be less powerful
and more efﬁcient. Problem 4.10. As computed in the text, an ideal kitchen refrigerator could have a COP of about
Tc 255 K Th—TC = 298K—255K 2
Therefore, by the deﬁnition of COP, QC 2 5.9W or W = QC / 5.9. In each second, this refrigerator must remove 300 J of heat from the inside, so the work required is W : 300 J / 5.9 2 50 J. In other words, the power drawn from the wall could be as little as 50 W.
(In practice the operation won’t be ideal, of course.) COP = 5.9. ,J 0‘ .. uuwu Funny. Problem 4.15. First let me draw an energyﬂow diagram for the absorption refrigerator: Absorption
refrigerator This relation says nothing about the ratio Qc/Qf (either Qc or Q; could be bigger
than the other), so energy conservation does permit the GOP to be greater than 1. Q? Qt: Qf
__ > _ __
21'  Tc + Since the COP involves QC and Qf but not (2,, let’s use energy conservation to elimi
nate Q,:
a+a>a+e a_a>a a I; 1121. Tf’ °r 21 T; *f‘f'
Solving for Qc/Qf and being careful with the direction of the inequality, we ﬁnd Problem 4.16. Hook up the hypothetical engine to the Carnot refrigerator as shown below, so each uses the same reservoirs and the refrigerator uses all the work produced by
the engine: ' 3'7.H.0trss'érvoia.Th”fz 5:? Ith It " HYPothetical " 4 ‘5 i { Carnot
engine ‘  For the ideal Carnot refrigerator, the heat input and output are in the same ratio as the
reservoir temperatures: For a given W, this equality and energy conservation (QM — Qt,r = W) determine the
values of Q,” and Q”. If the hypothetical engine were ideal, the same equalities would
apply to it, so we would have QM, = QM. and che m Qw. However, if the hypothetical
engine is better than ideal, then it requires a smaller amount of heat input to produce
the same amount of work, so QM < QM. Furthermore, energy conservation dictates that
its waste heat output must be smaller by the same amount, so QC”, < chr. Thus, the
net elfect of the engineurefrigerator combination is to transfer heat (in an amount equal
to Q,” — QM) from the cold reservoir to the hot reservoir, with no work input. It is a
“perfect” refrigerator, too good to be true. We are therefore forced to conclude that no
such hypothetical engine could possibly exist. Problem 4.21. (Stirling engine.) (a) The cycle consists of two isothermal processes (at Th and Tc) connected by constant—
Volunie processes: Th isotherm Tc isotherm V1“ V2 (b) To calculate the efﬁciency we need to know the net work done and the total heat input
for one cycle. The work done by the gas during the power stroke is V V
2 2 Wm: pdV: N th=NkThlnE,
V1 V1 V Vi and the work done by the gas during the compression stroke is similarly V
lNch V2
W = =_N _
34 L V chlﬂm, so the net work done per cycle is V
W = W12 + W34 = Nk(Th — Tc) ln 1
Meanwhile the heat input occurs during the power stroke and during the transfer to
the hot cylinder. Because the power stroke is isothermal, the energy of the gas doesn’t change during this step and therefore, by the ﬁrst law, V
Q12 = W12 = NkTh In 72.
1
During the transfer to the hot cylinder there is no work done, so by the ﬁrst law and
the equipartition theorem, the heat input is Q41 2 U1— vi = £~Nk(Th — Tc), (c) (d) where f is the number of degrees of freedom per molecule. Thus the total heat input
is V
at = Q12 + Q41: Nice. 1n V2 + given a To).
1 Now the efﬁciency is deﬁned as e = W/Qh. It’s algebraically simpler to compute the
reciprocal, '1_%%_NkThan/E/l/ill'gNMTh—E) _ Th _,_ f
e ‘ W _ Nk(T,, — Tc)ln(V2/V1) _ Th — :1"c 2111(1/2/1/1)‘
The ﬁrst term is just the reciprocal of the Carnot eﬁiciency, ea 3 1 —~ (Tc/Th), so we
can write 1 1 f
a. : _ + e co 21n(V2/V1)' The second term is always positive, but is smaller for large compression ratios. There—
fore 1/e is always greater than 1 /ec, that is, e is less than so, by an amount that is
smaller when the compression ratio is large. For instance, if Th is twice as large as Tc
(say 600 K compared to 300 K), so so = 1 / 2, and if the compression ratio is 10 and
the gas has ﬁve degrees of freedom per molecule, then 1 1 5 e=ﬁ+21n10_ that is, the efﬁciency is about 32%. With an ideal regenerator, the heat input during step 41 comes for free, because it’s
exactly the same (gNMTh — 11)) as the heat output during step 23. Therefore only
Q12 should be counted as part of (2,, when computing the efﬁciency. Following the
steps of the preceding calculation, this means that the second term in the expression
for 1/e is no longer present, and therefore, e 2 cc. 5
2 —=2 1.1;
+22.3 + 3.1, To really obtain 8 2 ea, the temperature difference between the gas and whichever
reservoir it is exchanging heat with would always have to be inﬁnitesimal; therefore
the engine would have to operate inﬁnitely slowly, just like an ideal Carnot engine. To
get any power out of a Stirling engine you have to run it with nonzero temperature
differences, just like the Carnot engine considered in Problem 4.6. You might think,
then, that there is no advantage to a Stirling engine over a Carnot engine, but in fact
the Stirling engine turns out to be easier (mechanically) to build and operate, since
the walls of the cylinders can always be at the same temperature, and are always in
approximate thermal equilibrium with the gas that is near them. Both the Stirling
and Carnot engines should be contrasted with internal combustion engines, which
are, generally more compact (no external combustion chamber) but are limited to
certain types of fuel and are more polluting because the fuel never burns completely.
Compared to a steam engine, a Stirling engine has the advantage of greater simplicity
and possibly higher efficiency, but the disadvantage of (probably) lower power. At
present, it seems that some other type of engine is considered more practical than
the Stirling engine for virtually every application. Howaver, a Stirling engine can be
operated in reverse as a refrigerator, and I’ve read that Stirling refrigerators are quite
practical for liquefying small amounts of air. ...
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This note was uploaded on 04/15/2010 for the course PHYSICS 16 taught by Professor N.darnton during the Spring '10 term at Smith.
 Spring '10
 N.Darnton
 Physics

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